CHAP. XVI.] THE INVERTED ARCH. 335 



at any point, and the sum of the forces between that point and 

 A, is a maximum, at least the entire distance from A to P' 

 must be covered with the load. 



In Fig. 110 we have the weight P on the other side of the but 

 centre P'. The construction is identical with Figs. 109 and 108, 

 the position and the direction of action of the forces is now 

 different. Since the resultants A and P' now lie on the same 

 side of P, A and P' act in opposite directions, and since P' 

 must still act upwards, A must act downwards. In the neigh- 

 borhood of 5' the moment is zero. Between this point and B 

 the moments have the same signification as in Fig. 109 ; on 

 the other side the moments have then a different sign. In 

 order, then, that the moment at 5' shall be a maximum, the load 

 must cover the length from A to P, this last point being the 

 point at which a load causes no moment in 5'; for if any 

 point between A. and P were not loaded, as we have seen, a 

 load at that point would increase the moment at 5'. A load 

 beyond P, however, would diminish the moment at 5'. 



The above holds good for every point between A and P', and 

 therefore for the point of rupture or of maximum moment it- 

 self. In order that this maximum moment can be no more 

 increased, the load must extend from A beyond the centre to 

 that point at which a load being placed causes no moment at 

 the cross-section of rupture. 



As for the shearing force, at the end A it will evidently be 

 greatest for load from A to P', or over the half span, since 

 every load on the other side of P' diminishes this reaction. 

 Hence we have the following principles established : 



The moment at any cross-section of the stiffening truss is a 

 maximum, when the load reaches from the nearest end beyond 

 the centre to a point for which the moment at this cross-section 

 is zero. 



The above condition holds good,therefore, for the maximum 

 of all the maximum moments, or for the cross-section of rup- 

 ture itself. 



The maximum shearing force is at one end of the truss 

 when the adjacent half span is loaded. 



If the arc is unsymmetrical, we must understand by " half 

 span " the distance between the end and vertical through the 

 intersection of the outer arc ends produced. 



187. Example. As an illustration of the above principles, 



